Calculating upper and lower bound












0














Suppose there is a function
$ f(x)= 19n^2/5n +1-n $

I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2



or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.



So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?










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  • $dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
    – Yadati Kiran
    Nov 28 at 11:23












  • i can also assume $ f(n)=19n/5+1−n $ ??
    – Tom
    Nov 28 at 11:28












  • @YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
    – Tom
    Nov 28 at 11:30










  • Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
    – Yadati Kiran
    Nov 28 at 11:31
















0














Suppose there is a function
$ f(x)= 19n^2/5n +1-n $

I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2



or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.



So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?










share|cite|improve this question






















  • $dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
    – Yadati Kiran
    Nov 28 at 11:23












  • i can also assume $ f(n)=19n/5+1−n $ ??
    – Tom
    Nov 28 at 11:28












  • @YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
    – Tom
    Nov 28 at 11:30










  • Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
    – Yadati Kiran
    Nov 28 at 11:31














0












0








0







Suppose there is a function
$ f(x)= 19n^2/5n +1-n $

I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2



or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.



So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?










share|cite|improve this question













Suppose there is a function
$ f(x)= 19n^2/5n +1-n $

I want to calculate upper and lower bound. But I had this confusion that whether I have to calculate in terms of n^2? Because dominating term is n^2



or If I solve the above equation as $ f(x)= 19n/5 +1-n $ (canceling n in 1st term) then what? Then I have to calculate c1 and c2 in terms of n? because that will be the dominating term.



So, Kindly tell me that in which term do I have to calculate c1 and c2 i.e n or n^2?? Can I cancel n in $ 19n^2/5n $ or not? What will be its asymptotic form? $ f(n)∈Θ(n^2) $ or $ f(n)∈Θ(n) $ ?







functions asymptotics upper-lower-bounds






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asked Nov 28 at 11:07









Tom

31




31












  • $dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
    – Yadati Kiran
    Nov 28 at 11:23












  • i can also assume $ f(n)=19n/5+1−n $ ??
    – Tom
    Nov 28 at 11:28












  • @YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
    – Tom
    Nov 28 at 11:30










  • Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
    – Yadati Kiran
    Nov 28 at 11:31


















  • $dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
    – Yadati Kiran
    Nov 28 at 11:23












  • i can also assume $ f(n)=19n/5+1−n $ ??
    – Tom
    Nov 28 at 11:28












  • @YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
    – Tom
    Nov 28 at 11:30










  • Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
    – Yadati Kiran
    Nov 28 at 11:31
















$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 at 11:23






$dfrac{19n^2}{5n}=dfrac{19n}{5}$, you can use either assuming your function is $f(n)=dfrac{19n^2}{5n}+1-n$.
– Yadati Kiran
Nov 28 at 11:23














i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 at 11:28






i can also assume $ f(n)=19n/5+1−n $ ??
– Tom
Nov 28 at 11:28














@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 at 11:30




@YadatiKiran What will be its asymptotic form? f(n)∈Θ(n2) or f(n)∈Θ(n) ?
– Tom
Nov 28 at 11:30












Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 at 11:31




Yes. Its asymptotic form will be a function of $n$ in its least possible exponent.
– Yadati Kiran
Nov 28 at 11:31















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